The energy scale of the qubit eigenvalues in D-Wave machine is GHZ = 10^{-7} Hartree = 10^{-5} eV. This means that in our calculation 1 energy unit correponds to 10^{-5} eV or 100 mK.
Caculate the wave length of a phonon with energy 10^{-5} eV in the following way: The Debye energy is roughly 10^{-2} eV. This corresponds to a wave vector k=\pi /a = 1, where a is the lattice constant. The acoustic phonon has a linear dispersion relation: \omega = kv, where v is the velocity of sound. Inserting \omega = 10^{-2} and k=1 yields v=10^{-2}. Therefore k at an phonon energy of 10^{-5} eV correponds to k=10^{-3}. The wave length is \lambda = 2\pi /k \approx 12\times 10^3bohr=10^4 Angstrom which is perhaps of the order of the width of the Josephson junction. If instead we would require a much smaller phonon wave length of e.g. 10 Angstrom, then k=1 and the phonon energy would be \omega = 10^{-2} eV. This is very large and phonons of this wave length cannot be excited at a temperature of 100 mK.
D-wave operates below 100 mK and therefore phonons with a wave length shorter than 10^4 Angstrom cannot be excited. Are the Mooij-Delft experiments an indication that phonons of this wave length are nevertheless switching qubits?
When coupling to the environment the eigenvalues obtain an energetic width of 10^{-4} GHZ= 10^{-9} eV. If the environment would be phonons, they would have a wave length of 10^8 Angstron comprising 10000 qubits. But phonons of such long wave length can neither exist in the D-Wave machine, nor can they couple to a single qubit and stabilize it against phonon excitations of shorter wavelength.
The time in the plots correspond to seconds, if energy unit is pico eV = 10^{-12}
eV. If energy units are interpreted as 10^{-5} eV, then
10000 time units are 1 milli second
10 time units are 1 micro second
1 time unit is 100 nano seconds.